Test results - Codility

Tasks Details

easy

1. Fish

N voracious fish are moving along a river. Calculate how many fish are alive.

100%

You are given two non-empty arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.

The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.

Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:

0 represents a fish flowing upstream,

1 represents a fish flowing downstream.

If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:

If A[P] > A[Q] then P eats Q, and P will still be flowing downstream,

If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream.

We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.

For example, consider arrays A and B such that:

A[0] = 4 B[0] = 0 A[1] = 3 B[1] = 1 A[2] = 2 B[2] = 0 A[3] = 1 B[3] = 0 A[4] = 5 B[4] = 0

Initially all the fish are alive and all except fish number 1 are moving upstream. Fish number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, number 0 and 4, never meet and therefore stay alive.

Write a function:

class Solution { public int solution(int[] A, int[] B); }

that, given two non-empty arrays A and B consisting of N integers, returns the number of fish that will stay alive.

For example, given the arrays shown above, the function should return 2, as explained above.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000];

each element of array A is an integer within the range [0..1,000,000,000];

each element of array B is an integer that can have one of the following values: 0, 1;

the elements of A are all distinct.

Solution

Programming language used Java 8

Time spent on task 7 minutes

Notes

not defined yet

Task timeline

Code: 08:33:35 UTC, java, verify, result: Failed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A, int[] B) {
        // write your code in Java SE 8
        int result = 0;
        int max = 0;
        Stack<Integer> downFishStack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            if (B[i] == 0) {
                //왼쪽으로 가는방향인데 오른쪽으로 가는 방향이 없을경우
                if (downFishStack.isEmpty()) {
                    result++;
                }
                //왼쪽으로 가는애인데 오른쪽으로 가는 방향이 있었을 경우
                else {
                    // 오른쪽방향의 max값보다 왼쪽방향의 값이 클경우 오른쪽 방향 값들은 모두 삭제
                    while(!downFishStack.empty() || downFishStack.peek() < A[i]){
                        downFishStack.pop();
                    }
                }
            } else if (B[i] == 1) {
                downFishStack.push(A[i]);
            }
        }

        return result + downFishStack.size();
    }
}

Analysis

▶

example
example test

✘

RUNTIME ERROR
tested program terminated with exit code 1

Code: 08:35:01 UTC, java, verify, result: Failed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public int solution(int[] A, int[] B) {
        // write your code in Java SE 8
        int result = 0;
        int max = 0;
        Stack<Integer> downFishStack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            if (B[i] == 0) {
                //왼쪽으로 가는방향인데 오른쪽으로 가는 방향이 없을경우
                if (downFishStack.isEmpty()) {
                    result++;
                }
                //왼쪽으로 가는애인데 오른쪽으로 가는 방향이 있었을 경우
                else {
                    // 오른쪽방향의 max값보다 왼쪽방향의 값이 클경우 오른쪽 방향 값들은 모두 삭제
                    while(!downFishStack.empty() && downFishStack.peek() < A[i]){
                        downFishStack.pop();
                    }
                }
            } else if (B[i] == 1) {
                downFishStack.push(A[i]);
            }
        }

        return result + downFishStack.size();
    }
}

Analysis

▶

example
example test

✘

WRONG ANSWER
got 1 expected 2

Code: 08:40:02 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] A, int[] B) {
        // write your code in Java SE 8
        int result = 0;
        int max = 0;
        Stack<Integer> downFishStack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            if (B[i] == 0) {
                //왼쪽으로 가는방향인데 오른쪽으로 가는 방향이 없을경우
                if (downFishStack.isEmpty()) {
                    result++;
                }
                //왼쪽으로 가는애인데 오른쪽으로 가는 방향이 있었을 경우
                else {
                    // 오른쪽방향의 max값보다 왼쪽방향의 값이 클경우 오른쪽 방향 값들은 모두 삭제
                    while (!downFishStack.empty() && downFishStack.peek() < A[i]) {
                        downFishStack.pop();
                    }
                    if (downFishStack.empty()) {
                        result++;
                    }
                }
            } else if (B[i] == 1) {
                downFishStack.push(A[i]);
            }
        }

        return result + downFishStack.size();
    }
}

Analysis

▶

example
example test

✔

Code: 08:40:13 UTC, java, verify, result: Passed

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] A, int[] B) {
        // write your code in Java SE 8
        int result = 0;
        int max = 0;
        Stack<Integer> downFishStack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            if (B[i] == 0) {
                //왼쪽으로 가는방향인데 오른쪽으로 가는 방향이 없을경우
                if (downFishStack.isEmpty()) {
                    result++;
                }
                //왼쪽으로 가는애인데 오른쪽으로 가는 방향이 있었을 경우
                else {
                    // 오른쪽방향의 max값보다 왼쪽방향의 값이 클경우 오른쪽 방향 값들은 모두 삭제
                    while (!downFishStack.empty() && downFishStack.peek() < A[i]) {
                        downFishStack.pop();
                    }
                    if (downFishStack.empty()) {
                        result++;
                    }
                }
            } else if (B[i] == 1) {
                downFishStack.push(A[i]);
            }
        }

        return result + downFishStack.size();
    }
}

Analysis

▶

example
example test

✔

Code: 08:40:17 UTC, java, final, score: 100

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

// you can also use imports, for example:
import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

class Solution {
    public static int solution(int[] A, int[] B) {
        // write your code in Java SE 8
        int result = 0;
        int max = 0;
        Stack<Integer> downFishStack = new Stack<>();
        for (int i = 0; i < A.length; i++) {
            if (B[i] == 0) {
                //왼쪽으로 가는방향인데 오른쪽으로 가는 방향이 없을경우
                if (downFishStack.isEmpty()) {
                    result++;
                }
                //왼쪽으로 가는애인데 오른쪽으로 가는 방향이 있었을 경우
                else {
                    // 오른쪽방향의 max값보다 왼쪽방향의 값이 클경우 오른쪽 방향 값들은 모두 삭제
                    while (!downFishStack.empty() && downFishStack.peek() < A[i]) {
                        downFishStack.pop();
                    }
                    if (downFishStack.empty()) {
                        result++;
                    }
                }
            } else if (B[i] == 1) {
                downFishStack.push(A[i]);
            }
        }

        return result + downFishStack.size();
    }
}

Analysis summary

The solution obtained perfect score.

Analysis

Detected time complexity:

O(N)

▶

example
example test

✔

▶

extreme_small
1 or 2 fishes

✔

0.004 s

0.008 s

0.004 s

▶

simple1
simple test

✔

▶

simple2
simple test

✔

▶

small_random
small random test, N = ~100

✔

▶

medium_random
small medium test, N = ~5,000

✔

▶

large_random
large random test, N = ~100,000

✔

▶

extreme_range1
all except one fish flowing in the same direction

✔

▶

extreme_range2
all fish flowing in the same direction

✔